Problem: Find $\lim_{h\to 0}\dfrac{2\tan\left(\dfrac{\pi}{3}+h\right)-2\tan\left(\dfrac{\pi}{3}\right)}{h}$. Choose 1 answer: Choose 1 answer: (Choice A) A $1$ (Choice B) B $2$ (Choice C) C $8$ (Choice D) D The limit doesn't exist
Answer: The limit expression has the following form: $\lim_{h\to 0}\dfrac{f(t+h)-f(t)}{h}$ Therefore, it describes a derivative of a certain function $f$ at a certain $x$ -value $t$. Can you recognize what $f$ and $t$ are? Since the numerator is $2\tan\left(\dfrac{\pi}{3}+h\right)-2\tan\left(\dfrac{\pi}{3}\right)$, we can tell that the function is $f(x)=2\tan(x)$ and the $x$ -value is $\dfrac{\pi}{3}$. In other words, the limit expression is equal to $f'\left(\dfrac{\pi}{3}\right)$ for $f(x)=2\tan(x)$. Let's find $f'(x)$ : $f'(x)=2 \cdot \sec^2(x)$ Now let's evaluate $f'\left(\dfrac{\pi}{3}\right)$ : $\begin{aligned}f'\left(\dfrac{\pi}{3}\right)&=2 \sec^2\left(\dfrac{\pi}{3}\right) \\\\ &=2\cdot (2)^2 \\\\ &=8\end{aligned}$ In conclusion, $\lim_{h\to 0}\dfrac{2\tan\left(\dfrac{\pi}{3}+h\right)-2\tan\left(\dfrac{\pi}{3}\right)}{h}=8$.